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Rough Approximation The angles between final trajectories are small enough that we could approximate this as a 1D problem. Let's assume both vehicles decelerated at $a$ after impact. $$d_A=\frac{{V_A}^2}{2a}$$ $$d_B=\frac{{V_B}^2}{2a}$$ The final momentum is then: $$p=M_AV_A+M_BV_B=M_A\sqrt{2d_Aa}+M_B\sqrt{2d_Ba}$$ Conserving momentum: ...


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When you jump from any given height, the force pulling you down is gravity with $F=mg$. This makes you accelerate to faster speeds as you fall farther, obviously. When you hit the ground, you do not experience the same acceleration. Otherwise, it would take just as long to stop falling as it took to get up to that speed. Hitting the ground imparts a much ...


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You're looking at the net force acting on him at the instant he begins falling. And you are correct that, in the absence of air resistance, the net force that is acting on the man the instant after he begins falling in the same, given by $F_g=mg$. However, there's an old saying that says, "it's not the fall that kills you; it's the sudden stop at the end." ...


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Recall that force is equivalent to, $$F=\frac{dp}{dt} \sim \frac{\Delta p}{\Delta t}$$ Where $p$ is the momentum, and $t$ is time. Momentum is given by, $$p=m \cdot v$$ Where $m$ is mass, and $v$ is velocity. When you hit the ground falling from 2 meters. The change in velocity is very small, and the change in time is small. When you hit the ground ...


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Suppose in zero gravity you squirt a smooth stream of water from a nozzle. A cylindrical tube of water with surface tension is not a minimum energy configuration. Spherical drops are minimum energy. The cylindrical tube is a metastable state, like a ball perfectly balanced on a point. Any slight variation from a perfect cylinder will be increasingly ...


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There's a good way to get an estimate out of this. First, perform dimensional analysis to get an estimate. Start with, finding ${{\Delta a} \over {\Delta v}}$, which is the change in the drag acceleration with respect to velocity. The argument that follows is physical rather than technical, so beware. The above expression, should be proportional to the ...


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I reckon I'm going to get lit up like RoboCop because of this post but here we go: I think the apparent separation of the water into droplets the higher up you go is caused because of two reasons: gravity and the cohesion/adhesion of water (also see Van der Waal forces). Essentially, the cohesive/adhesive forces are responsible for keeping the water ...


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If the door has only air drag acting on it, its acceleration is $a=-\beta v^2$ where $\beta$ is some drag coefficient and $v$ is speed. The time and distance needed to slow down from $v_1$ to $v_2$ is defined by $$ \begin{aligned} \Delta t & = \int \limits_{v_1}^{v_2} \frac{1}{a}\,\mathrm{d}v = \frac{1}{\beta} \left( \frac{1}{v_2} - \frac{1}{v_1} ...


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Once the door is detached from the front car, it will start decelerating, because of air drag, friction with the floor, etc. You can start assuming that this decelaration is constant, it is not perfectly accurate, as the door will likely be rotating, and being an irregular form, the drag will vary chaotically. Also the drag vary with velocity. If you assume ...


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The door is not slowing down because of its mere detachment from the front car. It is slowing down because of air drag on the door. That drag would depend on the area of cross-section interaction, the mass of the door and whether the door is tumbling. One might be able to estimate a worst-case scenario, but there isn't really a way to calculate an ...


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There is no such thing as classical motion of an electron in an atom. The quantum states electrons in an atom are in are atomic orbitals, which possess a definite energy, but not a definite position. The Bohr model of the electron, in which electrons are thought of as classical particles orbiting the nucleus, is false. The question whether or not two ...


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Why don't electrons collide among themselves Because they aren't anything like billiard balls. Check out the wave nature of matter. And take a look at the Wikipedia atomic orbitals article: "The electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves". The Heisenberg principle states that we ...


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Let the balls meet at the height $d$ from the ground then at the time of collision, the velocity of the ball (mass $m$) is given by third equation of motion as follows $$v^2=(0)^2+2g(h-d)$$ $$v^2=2g(h-d)\tag 1$$ & the time taken by ball (mass $m$) to reach at the point of collision is given by second equation of motion as follows ...


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Let $t$ be the time from the dropping the fist ball until the collision of the balls. Then, $v_1=gt$ and $v_2=u-gt$. Moreover, $d=ut-\frac{1}{2}gt^2$ and $h-d=\frac{1}{2}gt^2$, so that $ut=d+(h-d)$, which gives $t=\frac{h}{u}$. Because $mv_1=m_1v_2$, we must have $m\left(\frac{gh}{u}\right)=m_1\left(u-\frac{gh}{u}\right)$, or ...


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As a distant observer we can watch the shadow of the black holes forming in front of the background stars. According to a nice little paper by Daisuke Nitta, Takeshi Chiba, and Naoshi Sugiyama ("Shadows of Colliding Black Holes, 2011") the answer is yes. To a distant observer in a finite span of time two black holes form a shadow that is indistinguishable ...


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What is missing: this must be recognized as a relativistic process and be treated accordingly. Why is it a relativistic process: energy transforms into mass or conversely. This is something that cannot be accounted for using only classical mechanics. In older times I would've added "because it needs to account also for relativistic mass", but since using ...


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Two photons moving in opposite directions ("head-on") can collide and move off in different directions (still opposite if the photons have equal energies), If they have enough energy, the photons might produce an electron-positron pair. At even higher energies, other final states are allowed by conservation of energy. The cross-sections (or probabilities) ...


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Usually absolutely nothing. Electromagnetism is linear, which means that the result of doing something with two photons is the superposition of the results of doing something with each one individually. By that reasoning, since one photon by itself just goes on its own merry way, then two photons, even if they go near each other, just go along on their ...


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To leading order, nothing happens in any photon "collision". To higher order there are light-light interactions that involve particle loops, but they don't (can't) depend on the geometry because we can always boost to a frame in which the pair has zero net momentum (even though you can't boost to the frame where a single photon has zero momentum).


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Two gamma ray photons are produced by the collision of an electron and a positron. If the laws of physics are reversible at the quantum level, there is a chance that the collision of two gamma ray photons could produce a positron and an electron. However, it's questionable that this would not violate the 2nd law of thermodynamics. It seems to me that a ...


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Photons don't directly interact with each other, but if one photon pair produced an e+/e- then the second photon could interact with that pair. The interaction has to conserve the energy of the two photons and conserve their momentum as well of course. But yes they could (and most probably depending on their energy) just pass right "through" each other.


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Andromeda and the Milky Way belong to a group of galaxies called the Local Group. The two galaxies are the largest galaxies in the group, so to a pretty good approximation their interaction can be treated as a two body problem, with the other galaxies in the group producing only minor perturbations to their motion. So as you suspected, it isn't the case ...


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Their mutual gravity will pull them towards each other, with the more massive galaxy causing more acceleration on the smaller. According to this article the more massive is the Milky Way, so it will cause Andromeda to accelerate more than the Milky Way, not that it really matters. Since interstellar space is mostly empty (i.e. there is a lot of distance ...


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Strictly speaking, the universe has no rest frame (that we know of). If you want to set the rest frame of the CMB to be stationary, (which is reasonable in many applications), you may compare the velocity of the Milky Way and the velocity of Andromeda with respect to the CMB rest frame. (I don't know if anyone has ever done this, however.) The CMB dipole ...


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In an inelastic collision, some of the energy is absorbed by the colliding bodies - this is why you cannot use conservation of energy to calculate the resultant velocities of the bodies involved - you don't know how much is absorbed. But you do know that momentum is conserved, and assuming that the bodies remain intact (no pieces are separated from the ...


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Solving this particular set of equations would yield two values for $u1$ and as such two values for $u2$ - 10 and 5 ,that is, if the value of $u1$ were 5 the value of $u2$ would be 10 and vice-versa.You may arrive at this by solving a quadratic equation in $u1$ (or for that matter,$u2$) Now,you may have, while solving quadratics seen we,sometimes acquire ...


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Suppose someone suggests that following a perfectly elastic collision, two billiard balls are each traveling twice as fast as they were before (and opposite to their original directions). You can't prove him wrong using conservation of momentum, but you can prove him wrong using conservation of energy. Therefore conservation of energy has implications that ...


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The answer is that there is no simple answer. The way that energy and momentum get split up in the aftermath of a collision depends on the details of the collision itself, and there is nothing in the conservation laws themselves that influences this. The simplest case is in one-dimensional collisions, where both objects are constrained to move along the ...


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Isaac Newton observed the actions and reactions of objects in motion and recorded his observations as three famous laws. These laws (plus the conservation of momentum and of energy) can be used to explain how momentum and velocity are distributed among the objects coming out of a collision: (1) An object in motion will remain in motion with the same speed ...



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